2HI -> H2 I2
Keq= 49.5 at 440'C
If 5 0. M HI is initially placed into a container, what will be the equilibrium [ ] HI ?
"K_{eq}=\\frac{[H_2][I_2]}{[HI]^2}"
"2HI \\to H_2+I_2\\\\"
"5-x" "\\frac{x}{2}" "\\frac{x}{2}"
The initian concentration is "5 M."
"K_{eq}=49.5" "=\\frac{\\frac{x}{2}\\times\\frac{x}{2}}{(5-x)^2}"
"\\implies197x^2-1980x+4950=0"
"\\implies x=4.68"
The equilibrium concentration of "[HI]" is "=5-x=5-4.68=0.32\\ M"
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